Xbz grouping question B Geely digital digit DP entry

B Geely digital
Time Limit: 1 s

[Description]
Bigboyouyun recently came to a magical conclusion. If a number can be divided by 10, it is called the auspicious number. Calculate the number of Geely numbers in a certain interval.

[Input]
The number of first behavior samples is N. In the next n rows, each row represents an input sample. Each input sample has two numbers, representing the start point A and end point B of a specific interval.
Note that the range is [a, B], 1 <= A <= B <= 10 ^ 9

[Output]
N rows. For the X input sample, enter the result of the sample in Line X.

[Input example]
2
1 10
1 20

[Output example]
0
1

[Hint] There is no lucky number in 1-10, and only 19 lucky numbers in 1-20
Bytes ------------------------------------------------------------------------------------

Digital DP beginners.

I started by referring to the non-recursive method of the digital DP getting started PPT. I typed a table and then calculated it. After half a day of YY, I still got the error t ^ t.

 

Here is a template for writing digital DP using DFS:

Http://www.cnblogs.com/jffifa/archive/2012/08/17/2644847.html

Code:

 1 #include <stdio.h> 2 #include <string.h> 3 int dp[100][100]; 4 int dig[100]; 5  6 int dfs(int i,int s,bool e) 7 { 8     if (!i) return (s==0?1:0); 9     if ((!e)&&(dp[i][s]!=-1))    return dp[i][s];10     int res=0;11     int u=e?dig[i]:9;12     for (int q=0;q<=u;q++)13         res+=dfs(i-1,(s+q)%10,e&&q==u);14     if (!e)    dp[i][s]=res;15     return res;16 }17 18 int f(int n)19 {20     int len = 0;21     while(n)22     {23         dig[++len] = n % 10;24         n /= 10;25     }26     return dfs(len,0,true);27 }28 29 int main()30 {31     int a,b,T;32     memset(dp,-1,sizeof(dp));33     scanf("%d",&T);34     while (T--)35     {36         scanf("%d %d",&a,&b);37         printf("%d\n",f(b)-f(a-1));38     }39     return 0;40 }

View code

 

Core: (from neopenx, orz)

 1 int dfs(int len,int remain,bool fp) 2 { 3     if(!len) return remain==0?1:0; 4     if(!fp&&dp[len][remain]!=-1) 5         return dp[len][remain]; 6     int ret=0,fpmax=fp?digit[len]:9; 7     for(int i=0;i<=fpmax;i++) 8         ret+=dfs(len-1,(remain+i)%10,fp&&i==fpmax); 9     if(!fp) dp[len][remain]=ret;10     return ret;11 }12 13 int f(int n)14 {15     int len=0;16     while(n)17     {18         digit[++len]=n%10;19         n/=10;20     }21     return dfs(len,0,true);22 }

 

Len: the number of digits currently scanned, from high to status

Remain: indicates the current status. As shown in this question, from the highest digit to the current digit, numbers and % 10

FP: This is also called "First Place". neopenx's understanding is the first time to create a table.

The basic idea of this DFS function is to write down a table and use

Fpmax: current highest bit

 

 

Extension: hdu2089

Http://blog.csdn.net/dgq8211/article/details/9296953

 

Xbz grouping question B Geely digital digit DP entry